Given that $A,B$ are ordinals, prove that $A=B$ or $A \in B$ or $B \in A$
My attempt
Assume $A \not = B$ and $B \not\in A$ and i want to prove that $A \in B$
Let $C = A \cap B$. I think if I prove $A = C$ I am done.
$C \subset A$ because $ C = A \cap B$ , but I am stuck on the other side.
Proving that $A \subset C$ ?!
I think I need to use the fact that in any non-empty subset of ordinals there is a minimum)
Assume $A\notin B$, $B\notin A$ and $A\neq B$. By symmetry we can assume that there is $C_1\in B\setminus A$.
Then $C_1\subset B$, $A\notin C_1$, $C_1\notin A$ and $C_1\neq A$.
Repeat the same argument with $A$ and $C_1$ to get a $C_2\in$ either $A\setminus C_1$ or $C_1\setminus A$.
You find an infinite sequence of $C_i$ belonging to either $B$ or $A$. This is impossible due to the well-ordering. Therefore $A\notin B$, $B\notin A$ and $A\neq B$ is false.